If the function $$f(x) = \frac{\sin 3x + \alpha \sin x - \beta \cos 3x}{x^3}, x \in \mathbb{R}$$, is continuous at $$x = 0$$, then $$f(0)$$ is equal to :

Limit Existence: For the limit to exist, the numerator must be 0 at $$x=0$$.

$$\sin(0) + \alpha \sin(0) - \beta \cos(0) = 0 \implies 0 + 0 - \beta(1) = 0 \implies \mathbf{\beta = 0}$$.

 Use Taylor Series:

$$f(x) = \frac{(3x - \frac{(3x)^3}{6} + \dots) + \alpha(x - \frac{x^3}{6} + \dots)}{x^3}$$

$$f(x) = \frac{(3+\alpha)x + (-\frac{27}{6} - \frac{\alpha}{6})x^3 + \dots}{x^3}$$

Coefficient of $$x$$ must be 0:

$$3 + \alpha = 0 \implies \mathbf{\alpha = -3}$$.

$$f(0) = \lim_{x \to 0} \frac{-\frac{27}{6}x^3 - \frac{(-3)}{6}x^3}{x^3} = -\frac{27}{6} + \frac{3}{6} = -\frac{24}{6} = -4$$.

Correct Option: D (-4)